Differential equation describing the flowing of water out of a tank

I have the following differential equation describing water flowing out of the bottom of a tank of uniform cross-sectional area under the action of gravity.

$$ {{dh} \over {dt}} = - K\sqrt h $$

where $h(t)$ is the water depth at time $t$ with $h_0$ being the initial depth. $K > 0$ is a constant.

I have no idea how to solve this. I only know how to separate variables to solve ODEs. How would I solve this one? This is one of the starred questions in the book. I can't find another example like this.

Solution 1:

You can still solve this with separation of variables. divide both sides by $\sqrt{h}$ and you have a separated equation.